Introduces average rate of change as the slope of a secant line, computed from a table, a graph or a formula, with units read off the fraction. Builds the difference quotient as the same slope with the gap named h, then supplies the vocabulary for describing a graph precisely: increasing and decreasing intervals, local extrema, and absolute extrema, including why the endpoints of a closed window matter.
Subject: Precalculus · 65 slides · symbolic lesson
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Title
Precalculus · Chapter 1 — Functions
§1.3 Rates of Change and Behavior of Graphs, pp. 67-83
Objectives
Five things, each one you can check yourself on paper.
OpenStax, Precalculus, §1.3 Rates of Change and Behavior of Graphs §1.3, pp. 67-83 — the pages these objectives are drawn from
Warm-up
A car travels 120 miles in 2 hours. Almost everyone can produce the number 60. The question is what that number is actually claiming.
Discussion prompt
Does saying the car averaged 60 miles per hour mean it was ever travelling at 60 miles per hour? Explain.
Hint: Could the car have done the whole trip at two different speeds?
Answer:
Not necessarily. The car might have driven the first hour at 40 and the second at 80, never once passing through 60 on a steady stretch, and the average would still be 60.
The average tells you the overall change divided by the overall time, which is a statement about the two endpoints and nothing in between. The whole journey has been replaced by one straight line.
That is exactly what makes it an average rather than a speed, and it is exactly what Chapter 12 fixes by shrinking the interval until there is nothing left in between to average over.
Concept
The average rate of change of a function between two inputs is the change in output divided by the change in input. Geometrically that is the slope of the straight line joining the two points on the graph.
average rate of change — For a function f and inputs a and b, the quotient of the change in output by the change in input: f(b) minus f(a), all divided by b minus a. It equals the slope of the secant line through the two corresponding points on the graph.
\[ \text{average rate of change} = \frac{\Delta y}{\Delta x} = \frac{f(b)-f(a)}{b-a} \]
Nothing about this is new arithmetic — it is the slope formula from Algebra 1, applied to two points that happen to lie on a curve rather than a line. What is new is the interpretation: because the curve bends between them, the number describes the trip overall and not the steepness at any particular moment.
Figure (svg): A curve with two marked points joined by a straight secant line, with the horizontal run and the vertical rise drawn as a right triangle beneath it, showing that the average rate of change is the slope of that line
OpenStax, Precalculus, §1.3 Rates of Change and Behavior of Graphs §1.3, pp. 67-70
Section
Section 1
Concept
Subtract the outputs, subtract the inputs in the same order, and divide. The order matters only in that it must be consistent, because reversing both signs leaves the quotient unchanged.
That last bullet is worth pausing on. A ball thrown up and caught at the same height has an average rate of change of zero over the whole flight, despite having moved a great deal. The average sees only the two ends.
Figure (svg): A curve with two marked points joined by a straight secant line, with the horizontal run and the vertical rise drawn as a right triangle beneath it, showing that the average rate of change is the slope of that line
OpenStax, Precalculus, §1.3 Rates of Change and Behavior of Graphs §1.3, pp. 67-71
Picture it
The rise and run drawn under the secant are exactly the numerator and denominator of the formula.
Figure (svg): A curve with two marked points joined by a straight secant line, with the horizontal run and the vertical rise drawn as a right triangle beneath it, showing that the average rate of change is the slope of that line
The curve wanders above the secant between the two points, which is the visual statement that the average is not the instantaneous rate anywhere in particular.
Worked example
Two inputs, two outputs, one division.
\[ \text{Find the average rate of change of } f(x)=x^2-3x \text{ from } x=1 \text{ to } x=4. \]
Evaluate at the first input
Why: One minus three.
\[ f(1) = -2 \]
Evaluate at the second
Why: Sixteen minus twelve.
\[ f(4) = 4 \]
Subtract in matching order
Why: Outputs on top, inputs underneath.
\[ \frac{4 - (-2)}{4 - 1} \]
Divide
Why: Six over three.
\[ = 2 \]
Figure (svg): The solution to Worked example from a formula shown as a ladder of expressions, one row per legal move
\[ \frac{f(4)-f(1)}{4-1} = \frac{4-(-2)}{3} = 2 \]
Verify: check the sign against the picture
Why: The function is higher at 4 than at 1, so it rose overall and the rate must be positive, which it is. Reversing both subtractions would give negative 6 over negative 3, which is still 2 — consistent order is what matters, not which end comes first.
OpenStax, Precalculus, §1.3 Rates of Change and Behavior of Graphs §1.3, pp. 68-69
Prediction
A function is higher at the left endpoint of an interval than at the right.
Predict first
What is the sign of the average rate of change over that interval?
Correct: Negative, because the output fell overall.
Why: The numerator is the later output minus the earlier one, which is negative when the function ends lower than it started. The denominator is positive because the inputs increase. A negative over a positive is negative. What happens in between genuinely does not matter — the function could rise and fall wildly and the average would be unchanged.
Worked example
The petrol price in a town, recorded each year.
\[ \begin{array}{c|ccc} \text{year} & 2010 & 2013 & 2018 \\ \hline \text{price} & 2.20 & 3.40 & 2.90 \end{array} \]
Identify the two endpoints asked for
Why: The years 2010 and 2018; the middle row is not used.
\[ \text{endpoints } 2010\text{ and } 2018 \]
Subtract the outputs
Why: The later price minus the earlier.
\[ 2.90 - 2.20 = 0.70 \]
Subtract the inputs in the same order
Why: The later year minus the earlier.
\[ 2018 - 2010 = 8 \]
Divide and attach units
Why: Dollars on top, years underneath.
\[ 0.0875\text{ dollars per year} \]
Figure (svg): A card showing that an average rate of change carries units formed by dividing the output units by the input units, with three examples: miles per hour, dollars per item, and degrees per minute
\[ \frac{2.90-2.20}{2018-2010} = 0.0875 \text{ dollars per year} \]
Verify: notice what the middle year did not do
Why: The price rose to 3.40 and then fell, but the average over the whole period never saw that. The average from 2013 to 2018 is negative, and averaging over the whole span hides it. An average rate of change is only ever a statement about the two ends of the interval you chose.
OpenStax, Precalculus, §1.3 Rates of Change and Behavior of Graphs §1.3, pp. 70-71
Trap
\[ \frac{\Delta x}{\Delta y} = \frac{4-1}{4-(-2)} = \frac{1}{2} \]
Divide the change in input by the change in output
Why: Both differences are correct; only their positions have been swapped.
The answer of one half is reported as the average rate of change.
A rate is output per input, so the output change goes on top. The correct value is 2, not one half.
The units catch this instantly. The answer should be in output units per input unit; the wrong version comes out in input units per output unit — seconds per metre rather than metres per second, which is audibly not a speed.
Say the units out loud before reporting a rate. It takes two seconds and catches every inversion of this kind, including the ones where the numbers happen to look plausible.
Sorting
A rate is output per input. Decide which of each pair is the output.
Sort into buckets
Sort each quantity by where it belongs in the fraction.
Faded example
Find the average rate of change of the squaring rule from 2 to 6.
Fill in the blanks
\frac48 = \frac___}}___ = ___
Why: The outputs are 36 and 4, differing by 32, and the inputs differ by 4, giving 8. Worth noticing: 8 is the sum of the two inputs, and that is not a coincidence — for the squaring rule the average rate of change from a to b always simplifies to a plus b, which the difference quotient in the next section explains.
Estimation
A curve rises gently from a height of 3 at the input 0, then steeply to a height of 23 at the input 4.
Predict first
Roughly what is its average rate of change over that interval?
Correct: About 5.
Why: The output changed by 20 while the input changed by 4, so the average rate is about 5 output units per input unit. The second option is the numerator alone with the division forgotten, the third is the fraction inverted, and the fourth multiplies instead of dividing. Estimating the size first makes all three of those wrong answers visibly wrong.
Section
Section 2
Concept
Fix one input a and call the gap to the second input h. The average rate of change from a to a plus h is the difference quotient, and it is a formula in a and h rather than a single number.
\[ \frac{f(a+h)-f(a)}{h}, \qquad h \ne 0 \]
The reason for writing it this way is entirely forward-looking. Once the answer is a formula in h, you can ask what happens as h shrinks towards zero, and that limit is the derivative. This lesson stops one step short of that, but it builds the expression the limit will be taken of.
Figure (svg): The difference quotient shown as a secant between the point at a and the point at a plus h, with the run labelled h and the rise labelled f of a plus h minus f of a
OpenStax, Precalculus, §1.3 Rates of Change and Behavior of Graphs §1.3, pp. 71-75
Picture it
The picture is the same secant as before, with the second input written as a plus h instead of b.
Figure (svg): The difference quotient shown as a secant between the point at a and the point at a plus h, with the run labelled h and the rise labelled f of a plus h minus f of a
As h shrinks, the second point slides back along the curve towards the first and the secant pivots. Chapter 12 takes that process to its conclusion; here we only build the formula.
Worked example
Expand carefully, then watch every surviving term keep a factor of h.
\[ \text{Find the difference quotient for } f(x)=x^2. \]
Write f at the shifted input
Why: Substitute the whole expression, in brackets.
\[ (a + h) ^{2} \]
Expand it
Why: The square of a binomial, with its middle term.
\[ a ^{2} + 2 a h + h ^{2} \]
Subtract f(a) and simplify the numerator
Why: The a squared terms cancel.
\[ 2 a h + h ^{2} \]
Divide by h and cancel
Why: Every surviving term has a factor of h.
\[ 2 a + h \]
Figure (svg): The solution to Worked example the difference quotient of a quadratic shown as a ladder of expressions, one row per legal move
\[ \frac{(a+h)^2-a^2}{h} = \frac{2ah+h^2}{h} = 2a+h \]
Verify: test against the earlier computation
Why: Taking a equal to 2 and h equal to 4 gives 4 plus 4, which is 8 — exactly the average rate of change from 2 to 6 found in the previous section. And it explains the pattern noticed there: 2a plus h is the same as a plus the quantity a plus h, which is the sum of the two inputs.
OpenStax, Precalculus, §1.3 Rates of Change and Behavior of Graphs §1.3, pp. 72-73
Prediction
The difference quotient is formed for a linear rule with slope 5.
Predict first
What will it simplify to?
Correct: 5, with no h and no a remaining.
Why: A straight line has the same steepness everywhere, so every secant on it is the line itself and every average rate of change equals the slope. The quotient must therefore come out as the constant 5. This is the one case where an answer with no a in it is correct, and it is correct precisely because the function is linear.
Worked example
The reciprocal rule, where the cancellation is less obvious.
\[ \text{Find the difference quotient for } f(x)=\frac{1}{x}. \]
Write the numerator
Why: One over the shifted input, minus one over the original.
\[ \frac{1}{a + h} - \frac{1}{a} \]
Combine over a common denominator
Why: The common denominator is a times a plus h.
\[ \frac{a - (a + h)}{a(a + h)} \]
Simplify the top
Why: The a terms cancel, leaving negative h.
\[ -\frac{h}{a(a + h)} \]
Divide by h, which is multiplying by one over h
Why: The h cancels against the one in the numerator.
\[ -\frac{1}{a(a + h)} \]
Figure (svg): The solution to Worked example a difference quotient needing a common denominator shown as a ladder of expressions, one row per legal move
\[ \frac{\frac{1}{a+h}-\frac{1}{a}}{h} = \frac{-1}{a(a+h)} \]
Verify: check the sign makes sense
Why: The reciprocal rule is decreasing wherever it is defined, so every secant between two points on the same branch must slope downward and the quotient must be negative. For positive a and positive h the denominator is positive, so the whole expression is negative, as required.
OpenStax, Precalculus, §1.3 Rates of Change and Behavior of Graphs §1.3, pp. 74-75
Error analysis
A student forms the difference quotient of the squaring rule.
Annotate
On: \( \frac{(a+h)^2-a^2}{h} = \frac{a^2+h^2-a^2}{h} = \frac{h^2}{h} = h \)
A sanity check that costs nothing: the answer to a difference quotient should almost always still mention a, because a function's steepness usually varies from place to place. An answer free of a is claiming the function is a straight line.
Faded example
Forming the difference quotient for the rule that cubes its input, at the numerator stage.
Fill in the blanks
(a+h)^3 = a^3 + 3a^2h + 3ah^2 + h^3 \;\Longrightarrow\; \frach^2___ = 3a^2 + 3ah + ___
Why: The cube of a binomial has four terms with coefficients 1, 3, 3, 1 — a row of Pascal's triangle, which §11.6 will make systematic. After the a cubed terms cancel, every remaining term carries at least one h, so dividing by h leaves a polynomial with no fraction in it. Setting h to zero in that result would give three a squared, which is the derivative of the cubing rule.
Socratic
The difference quotient carries the condition that h is nonzero.
Discussion prompt
Explain why h cannot be zero, and why the answer after simplifying nevertheless makes perfect sense at h equal to zero.
Hint: What is the expression before simplifying, and what is it after?
Answer:
Before simplifying, the expression has h in the denominator, so h equal to zero is a division by zero and is meaningless. There is also no secant to speak of: both points would be the same point, and two identical points determine no line.
After simplifying, the h has cancelled and the result — 2a plus h, say — is a perfectly ordinary expression that can be evaluated at h equal to zero. But that value was not obtained by dividing by zero; it was obtained by simplifying first and substituting after.
That gap between the two is exactly what a limit formalises, and it is the reason Chapter 12 exists. The value the simplified expression takes at zero is the slope the secants are heading towards, and calling it the derivative is the whole of the next idea in the subject.
Matching
Each has been fully simplified.
Match the pairs
Why: The linear rule gives a constant, because a line has one slope everywhere. The others all still mention a, because their steepness varies. Setting h to zero in each gives 2a, 3, negative one over a squared, and three a squared, which are exactly the derivatives Chapter 12 will define — so this table is a preview of the derivative rules with the limit not yet taken.
Section
Section 3
Concept
A function is increasing on an interval when larger inputs there give larger outputs, and decreasing when larger inputs give smaller outputs. These are statements about stretches of the input axis.
Naming intervals with outputs is the standard error, and it produces answers that are not merely wrong but nonsensical: saying a function increases 'on the interval from 2 to 7' when 2 and 7 are heights describes no stretch of the domain at all.
Figure (svg): A curve rising, then falling, then rising again, with the rising stretches shaded green and the falling stretch shaded red, and the two turning points marked
OpenStax, Precalculus, §1.3 Rates of Change and Behavior of Graphs §1.3, pp. 76-79
Picture it
The shading marks stretches of the input axis, which is where this vocabulary lives.
Figure (svg): A curve rising, then falling, then rising again, with the rising stretches shaded green and the falling stretch shaded red, and the two turning points marked
Each turning point separates two shaded bands and belongs to neither. Reading the shading onto the horizontal axis gives the intervals directly.
Worked example
Using the curve on the visual slide, which turns at the inputs negative 1 and 2.
\[ \text{State where the graph is increasing and where it is decreasing.} \]
Start from the far left and read rightwards
Why: The curve climbs until the first turn.
\[ \text{rising until } x = -1 \]
Note the first turning point
Why: It peaks and starts down.
\[ \text{turn at } x = -1 \]
Continue to the second turn
Why: It falls through to a trough.
\[ \text{falling from } -1\text{ to } 2 \]
Finish to the right
Why: It climbs again and does not turn back.
\[ \text{rising after } x = 2 \]
Figure (svg): The solution to Worked example read the intervals off a graph shown as a ladder of expressions, one row per legal move
\[ \text{increasing on } (-\infty,-1)\cup(2,\infty), \qquad \text{decreasing on } (-1,2) \]
Verify: check the intervals are named with inputs
Why: Every number written down — negative 1, 2, and the infinities — is a value on the horizontal axis. If any of the heights of the turning points had appeared in the answer, that would be the standard error. Note also that the increasing set is a union of two separate stretches, which is normal and not a mistake.
OpenStax, Precalculus, §1.3 Rates of Change and Behavior of Graphs §1.3, pp. 77-78
Discrimination
Each statement uses a number. Decide which axis that number lives on.
Sort into buckets
Sort each statement by what its numbers describe.
Worked example
Average rates of change over short intervals reveal the behaviour without a graph.
\[ \begin{array}{c|ccccc} x & 0 & 1 & 2 & 3 & 4 \\ \hline f(x) & 5 & 8 & 9 & 7 & 2 \end{array} \]
Find the change across each step
Why: Subtract consecutive outputs.
\[ +3, +1, -2, -5 \]
Interpret the positive changes
Why: The function rose across the first two steps.
\[ \text{rising on } 0\text{ to } 2 \]
Interpret the negative changes
Why: It fell across the last two.
\[ \text{falling on } 2\text{ to } 4 \]
Locate the turn between them
Why: The sign of the change flips between the second and third step.
\[ \text{turns near } x = 2 \]
Figure (svg): The solution to Worked example decide from a rate of change shown as a ladder of expressions, one row per legal move
\[ \text{increasing on } (0,2), \quad \text{decreasing on } (2,4) \]
Verify: state the limits of what a table can show
Why: A table shows the behaviour only at the inputs listed. The function could dip and recover between 0 and 1 without the table noticing. So the honest conclusion is that it rises overall on the first stretch, and that the turning point is near 2 rather than exactly at it — a graph or the calculus would be needed to say more.
OpenStax, Precalculus, §1.3 Rates of Change and Behavior of Graphs §1.3, pp. 78-79
Trap
\[ \text{increasing on } (1.25, \; 7.25) \]
Read the heights at the two ends of the rising stretch
Why: The curve climbs from a height of 1.25 to a height of 7.25.
Those two heights are reported as the interval on which the function increases.
Intervals of increase are stretches of the input axis. The heights 1.25 and 7.25 are outputs, and quoting them describes no set of inputs at all.
What should be reported is the interval of inputs over which that climb happened — in the figure, from 1 to 5.
A quick test before writing the answer down: could you feed the numbers in your interval into the function? If not, they are outputs and belong in a statement about the range instead.
Prediction
Every average rate of change of a function over every subinterval of a stretch comes out positive.
Predict first
What can you conclude on that stretch?
Correct: It is increasing there.
Why: A positive average rate of change over an interval says the output ended higher than it started. If that holds over every subinterval, no matter how small, then for any two inputs in the stretch the later one gives the larger output, which is exactly the definition of increasing. The word 'every' is what makes the argument work — a single positive average over the whole stretch would not rule out a dip in the middle.
Two truths and a lie
Two of these are true of increasing and decreasing intervals and one is false.
Eliminate the wrong options
One of these claims is wrong.
Survives elimination: B
Why: B is the false claim. A turning point belongs to neither interval, which is why these intervals are conventionally written open. Including it in both would say the function is simultaneously rising and falling at that input, which is incoherent.
Explain it to yourself
Increasing and decreasing were defined without mentioning rates of change.
Discussion prompt
Explain the connection between the sign of an average rate of change and whether a function is increasing.
Hint: What does a positive numerator mean when the denominator is positive?
Answer:
Over an interval where the inputs increase, the denominator of the rate is positive. So the sign of the rate is the sign of the change in output: positive means the function ended higher, negative means lower.
Increasing on a stretch means this holds for every pair of inputs within it, so every average rate of change over every subinterval is positive.
This is why calculus can eventually decide the question mechanically. Shrink those intervals to zero and the sign of the resulting derivative decides increase or decrease at each individual point, without any need to look at a picture.
Section
Section 4
Concept
A local maximum is a point whose output is at least as large as those at every nearby input. It need not be the largest output the function ever produces.
local maximum — A point at which the output is greater than or equal to the outputs at all nearby inputs. The input where it happens is its location; the output there is its value. Also called a relative maximum.
That last point is the reason the next section has to exist separately. The endpoints of a closed interval can hold the largest and smallest values of a function without being local extrema at all, because there are no neighbours on one side to beat.
Figure (svg): A curve rising, then falling, then rising again, with the rising stretches shaded green and the falling stretch shaded red, and the two turning points marked
OpenStax, Precalculus, §1.3 Rates of Change and Behavior of Graphs §1.3, pp. 79-81
Picture it
The two marked points are where the shading changes colour, which is where the behaviour turns.
Figure (svg): A curve rising, then falling, then rising again, with the rising stretches shaded green and the falling stretch shaded red, and the two turning points marked
The peak on the left is a local maximum. Whether it is the highest point on the whole graph is a different question, and on this curve the answer is no.
Worked example
Two answers are wanted for each: where it happens, and what the value is.
\[ \text{From the graph, give the local extrema of } f. \]
Find where the curve stops rising
Why: The first turn, at the left peak.
\[ \text{local } \max\text{ at } x = -1 \]
Read the height there
Why: The output at that input.
\[ \text{value about } 6.7 \]
Find where it stops falling
Why: The trough between the two rises.
\[ \text{local } \min\text{ at } x = 2 \]
Read that height
Why: The output at the trough.
\[ \text{value about } -0.3 \]
Figure (svg): A curve rising, then falling, then rising again, with the rising stretches shaded green and the falling stretch shaded red, and the two turning points marked
\[ \text{local max at } x=-1, \quad \text{local min at } x=2 \]
Verify: say each answer in words
Why: There is a local maximum at the input negative 1, and its value is about 6.7. Getting these the wrong way round — reporting the location as 6.7 — is the standard error, and saying the sentence out loud makes it obvious, because 'a maximum at the input 6.7' would put it off the right-hand edge of the picture.
OpenStax, Precalculus, §1.3 Rates of Change and Behavior of Graphs §1.3, pp. 80-80
Sorting
A local extremum must beat its neighbours on both sides.
Sort into buckets
Sort each described point.
Worked example
Not every function has any, and recognising that is a real answer.
\[ \text{Find the local extrema of } f(x)=x^3. \]
Ask where it stops rising
Why: The cubing rule increases everywhere.
Check the flat spot at the origin
Why: The curve levels off there but does not turn.
Compare with nearby outputs
Why: Inputs just left of 0 give smaller outputs; just right, larger.
Conclude
Why: No point beats all of its neighbours.
Figure (svg): The solution to Worked example a function with no local extrema shown as a ladder of expressions, one row per legal move
\[ f(x)=x^3 \text{ has no local maximum and no local minimum.} \]
Verify: check the tempting wrong answer
Why: The origin looks special because the curve flattens there, and it is tempting to call it an extremum. But a maximum must be at least as high as its neighbours on both sides, and the outputs just to the right of 0 are larger. Flattening is not turning, and this distinction is exactly what the second derivative test in calculus is built to handle.
OpenStax, Precalculus, §1.3 Rates of Change and Behavior of Graphs §1.3, pp. 81-81
Error analysis
A student describes the left peak of the curve in this section.
Annotate
On: \( \text{local maximum at } x = 6.7 \)
Nearly every extremum question wants both numbers, and the marks are usually split between them. Write both, and label which is which.
Prediction
A polynomial's graph crosses the horizontal axis four times and is smooth throughout.
Predict first
At least how many local extrema must it have?
Correct: At least three, one between each pair of consecutive crossings.
Why: Between two consecutive crossings the curve leaves the axis, goes somewhere, and comes back, so it must turn at least once in between. Four crossings give three gaps and therefore at least three turns. Crossings themselves are not extrema — the curve is passing through, not turning — which rules out the second option.
Matching
Extremum questions come in two flavours and want different numbers.
Match the pairs
Why: The four questions alternate between the two axes, and telling them apart is most of what this section assesses. Note that 'where' and 'on what interval' both ask about inputs, while 'what value' and 'what range' both ask about outputs — the grammar is a reliable guide.
Counterexample
A classmate claims that wherever a graph flattens out, there is a local extremum.
Discussion prompt
Find a function and a point where the graph flattens and there is no extremum.
Hint: You have already met one in this section.
Answer:
The cubing rule at the origin. The curve levels off there — it is momentarily horizontal — but it continues to rise through, so outputs just to the right are larger and outputs just to the left are smaller.
So flattening is necessary but not sufficient for a smooth function to have an extremum. Every smooth extremum has a flat tangent; not every flat tangent is an extremum.
Calculus names this distinction: the flat points are the critical points, and a further test decides which of them are extrema. Meeting the counterexample now means that later test will look like an answer to a question you already have, rather than an extra rule.
Section
Section 5
Concept
An absolute maximum is the largest output the function attains anywhere on its domain. Unlike a local maximum, it must beat every other point, not merely its neighbours.
The practical procedure follows directly: to find the extreme values on a closed interval, list the outputs at every turning point and at both endpoints, then pick the largest and smallest from that list. Missing an endpoint is what makes an otherwise correct answer wrong.
Figure (svg): A curve on a closed window with a local maximum in the middle that is not the highest point overall, and a higher endpoint, showing the difference between a local and an absolute maximum
OpenStax, Precalculus, §1.3 Rates of Change and Behavior of Graphs §1.3, pp. 81-83
Picture it
Four points are marked, and only two of them are the extreme values of the function on this window.
Figure (svg): A curve on a closed window with a local maximum in the middle that is not the highest point overall, and a higher endpoint, showing the difference between a local and an absolute maximum
The absolute maximum sits at the right-hand endpoint, higher than the local maximum in the middle. Checking only the turning points would have found the wrong answer.
Worked example
Check the turning points and the endpoints, then compare the list.
\[ \text{Find the absolute extrema of the graphed } f \text{ on } [-3, 5]. \]
List the turning points inside the interval
Why: The local max and the local min.
\[ x = -1\text{ and } x = 2 \]
Add both endpoints to the list
Why: The interval is closed, so both are in the domain.
\[ x = -3\text{ and } x = 5 \]
Evaluate at all four
Why: Four candidate outputs.
\[ \text{about } 6.7, -0.3, -2.5, 6.7 - i s h \]
Pick the largest and the smallest
Why: The right endpoint wins on top; the left endpoint on the bottom.
\[ \max\text{ at } x = 5, \min\text{ at } x = -3 \]
Figure (svg): A curve on a closed window with a local maximum in the middle that is not the highest point overall, and a higher endpoint, showing the difference between a local and an absolute maximum
\[ \text{absolute max at } x=5, \qquad \text{absolute min at } x=-3 \]
Verify: notice that neither is a turning point
Why: Both extreme values happen at endpoints, and neither is a local extremum, because an endpoint has neighbours on one side only. A student who checked only where the curve turns would have reported the middle peak and trough and been wrong twice.
OpenStax, Precalculus, §1.3 Rates of Change and Behavior of Graphs §1.3, pp. 82-82
Sorting
The two kinds have different rules about endpoints.
Sort into buckets
Sort each statement.
Worked example
Unbounded domains can leave the question with no answer, and saying so is the answer.
\[ \text{Find the absolute extrema of } f(x)=x^2 \text{ on } (-\infty,\infty). \]
Find the turning point
Why: The parabola turns at its vertex.
\[ \text{vertex at } x = 0 \]
Evaluate there
Why: The output at the vertex.
\[ f(0) = 0 \]
Ask whether anything is lower
Why: A square is never negative.
\[ 0\text{ is the smallest output} \]
Ask whether anything is highest
Why: Outputs grow without bound in both directions.
Figure (svg): The solution to Worked example a function with no absolute maximum shown as a ladder of expressions, one row per legal move
\[ \text{absolute min } 0 \text{ at } x=0; \quad \text{no absolute maximum} \]
Verify: check why the guarantee does not apply
Why: The theorem that promises both extrema requires a closed and bounded interval, and the whole real line is neither. Restricting to the closed interval from negative 3 to 3 would restore the guarantee, and the maximum would then appear at both endpoints at once with the value 9 — a reminder that an absolute maximum value can be attained at more than one input.
OpenStax, Precalculus, §1.3 Rates of Change and Behavior of Graphs §1.3, pp. 83-83
Trap
\[ \text{turning points at } x=-1, \, x=2 \;\Longrightarrow\; \text{absolute max at } x=-1 \]
Find every place the curve turns and compare their heights
Why: The left peak is higher than the trough, so it is chosen as the maximum.
The middle peak is reported as the absolute maximum on the closed interval.
The endpoints are candidates too. On the interval shown, the right endpoint is higher than the middle peak, so it is the absolute maximum and the middle peak is only a local one.
An endpoint has neighbours on one side only, so it is never a local extremum on a smooth curve — but nothing stops it being the largest or smallest value on the interval.
The procedure is: turning points AND endpoints, then compare. This is the shape of every closed-interval optimisation problem in calculus, and forgetting the endpoints is the error that survives all the way into Chapter 12.
Prediction
A continuous function is considered on the open interval from 0 to 1, and it increases throughout.
Predict first
Does it attain an absolute maximum there?
Correct: No, because the right endpoint is excluded.
Why: The outputs climb towards the value the function would take at 1, but 1 is not in the interval, so that value is never attained. For any input in the interval there is a larger one still inside it, giving a larger output. This is exactly why the existence theorem insists the interval be closed — an open end is enough to break it.
Ranking
Each of these says more than the one before it.
Put in order
Why: Beating one nearby point is the weakest claim of the three. Beating all nearby points is a local maximum. Beating every point in the domain is an absolute maximum, and for an interior point it implies the local statement as well. The ordering makes clear why finding a local maximum does not finish an absolute-extremum question.
Edge cases
The theorem guarantees both extrema for a continuous function on a closed bounded interval.
Discussion prompt
Give an example showing the guarantee fails if the function is allowed to be discontinuous, even on a closed interval.
Hint: Try a piecewise rule that jumps at one point.
Answer:
Take the interval from 0 to 1 and the rule that gives x for every input below 1, and gives 0 at the input 1 itself. It is defined on a closed bounded interval, so the interval hypothesis holds.
Its outputs climb towards 1 but never reach it: the only input that could produce 1 is redefined to give 0 instead. So there is no absolute maximum, despite the closed interval.
Continuity is what rules this out, and this is why §12.3 spends a whole section on it. Each of the three hypotheses — closed, bounded, continuous — can be broken to produce a counterexample, which is the sign that none of them is decoration.
Comparison
Fill the blanks from memory. These two are the pair most often conflated in this chapter.
Comparison matrix
| local extremum | absolute extremum | |
|---|---|---|
| what it beats | nearby inputs only | every input in the domain |
| can be at an endpoint | no, on a smooth curve | yes, and often is |
| how many | possibly many | at most one value |
| guaranteed to exist | no | yes, if continuous on a closed bounded interval |
| found by | looking for turning points | comparing turning points and endpoints |
The second row is the one that costs marks: an endpoint can hold the absolute maximum without being a local maximum at all, so the two searches are not the same search.
Pattern
Given a graph and asked to describe its behaviour, this order gets everything and repeats nothing.
Step 4 is the one to build a habit around. On an unbounded domain it is skipped, but on a closed interval it decides the answer as often as not.
OpenStax Algebra and Trigonometry 2e, §3.3 Rates of Change and Behavior of Graphs §3.3
Check
Outputs on top.
Check your understanding
What is the average rate of change of f(x) = 2x squared from x = 1 to x = 3?
Answer: A
Why: The outputs are 2 and 18, differing by 16, and the inputs differ by 2, giving 8. Equivalently, the difference quotient for twice the squaring rule is 2 times the quantity 2a plus h, which at a equal to 1 and h equal to 2 gives 8.
Check
Expand carefully.
Check your understanding
The difference quotient for f(x) = x squared plus x simplifies to which expression?
Answer: A
Why: The squaring part contributes 2a plus h, as computed in this lesson, and the linear part contributes its slope of 1. Adding them gives 2a plus h plus 1. Difference quotients add across a sum of functions, which is worth noticing now because the derivative rules in Chapter 12 inherit that property.
Check
Remember the endpoints.
Check your understanding
A continuous function on the closed interval from 0 to 10 has one local maximum, at x = 4. Where can its absolute maximum be?
Answer: A
Why: The candidates for an absolute extremum on a closed interval are the turning points and the endpoints, so all three locations are possible and the outputs must be compared to decide. The theorem does guarantee existence here, since the function is continuous on a closed bounded interval.
Real world
Every reported growth figure is an average rate of change, and every one of them hides the middle.
Discussion prompt
A news report says a company's revenue grew at an average of 8 percent a year over the last decade. What does that figure guarantee, and what can it conceal?
Hint: How many of the ten years does the figure actually look at?
Answer:
It guarantees a relationship between the two endpoints — where revenue started and where it ended. That is genuinely all it claims.
It can conceal almost anything in between: a collapse in year three and a spectacular recovery in year nine produce the same average as steady growth. Averages over long intervals are exactly the tool for hiding volatility, which is why financial reporting also quotes year-on-year figures — those are the same calculation over much shorter intervals.
This is the same point as the warm-up about the car, and it is the reason calculus shrinks the interval. Shorter intervals hide less, and the limit of that process hides nothing at all.
Commit first
State your confidence along with your answer.
Predict first
A function has an average rate of change of zero over an interval. What must be true?
Correct: Its outputs at the two endpoints are equal.
Why: A rate of zero means the numerator vanished, so the two endpoint outputs match. Nothing else follows: the function may have climbed and fallen back, as the thrown ball did in the concept slide. Being constant would produce a zero rate but is far stronger than what was given, and a curve that rises and returns must indeed turn somewhere — but it could equally have fallen and returned, giving a local minimum instead.
Explain it
The test of understanding is being able to say why, not just what.
Discussion prompt
Explain to a classmate the difference between the average rate of change over an interval and the steepness of the curve at a single point, and why this lesson can only compute the first.
Hint: How many points does each one need?
Answer:
The average rate needs two points and reports the slope of the straight line joining them. It is a statement about a stretch.
Steepness at a single point would need the slope of a line touching the curve at just that point — but two points are needed to determine a line, and there is only one. That is the obstacle, and it is a real one rather than a gap in technique.
The difference quotient is the way around it: keep two points, but name the second relative to the first and watch what happens as the gap shrinks. This lesson builds the expression; Chapter 12 takes the limit. A good explanation makes clear that the obstacle is genuine, because that is what makes the eventual solution interesting.
Exit ticket
One honest answer, so the next lesson can start in the right place.
Predict first
Which idea from this lesson would you most want to see again?
Correct: Any of these is a legitimate answer; the useful one is the honest one.
Why: There is no correct choice here. The second is the most algebraically demanding and the one that pays off most directly in Chapter 12. The third and fourth cost marks for reasons that are about precision of language rather than difficulty, which makes them cheap to fix once noticed.
Connect it up
One page, drawn from memory, is worth more than rereading the section.
Draw it
Sketch a curve that has two local maxima, one local minimum, and an absolute minimum at its left endpoint on a closed interval. Label every one of those points with both its location and its value. Then draw a secant line between two points on your curve and write the fraction whose value is that secant's slope, using the letters a and a plus h.
If your absolute minimum sits at an endpoint and is not one of your local minima, you have drawn the distinction this lesson exists to make.
Recap
Five things, and the second one is the expression Chapter 12 takes a limit of.
| if you remember one thing | it should be this |
|---|---|
| about rates | a rate of change is the slope of a secant line |
| about the difference quotient | the h always cancels, and what is left still mentions a |
| about intervals | name them with inputs, never with outputs |
| about extrema | on a closed interval, check the endpoints too |
Section 1.4 stops looking at one function at a time and starts feeding one into another, which is how the rest of the course builds complicated rules out of the toolkit.
OpenStax, Precalculus, §1.3 Rates of Change and Behavior of Graphs §1.3, pp. 67-83 — everything on these slides traces back here
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